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ENGR5002 / Week 9
ENGR263 2019 Q A2Past paperOld spec ENGR2632:125 marks30 min

ENGR263 Process of Mass Transfer 2019 Q A2[VALID]

Answer ALL parts (a) - (b).

Formulas you may need
  • Schmidt number Sc=μρDAB=νDAB\mathrm{Sc} = \dfrac{\mu}{\rho D_{AB}} = \dfrac{\nu}{D_{AB}}, Sherwood number Sh=kcLDAB\mathrm{Sh} = \dfrac{k_cL}{D_{AB}} (on the formula sheet)
  • Diffusion through a stagnant gas: NA=PDABRTδln⁡P−pA2P−pA1N_A = \dfrac{PD_{AB}}{RT\delta}\ln\dfrac{P - p_{A2}}{P - p_{A1}} (on the formula sheet)
  • DAB∝T3/2/PD_{AB} \propto T^{3/2}/P (on the formula sheet)
  • Multicomponent stagnant mixture 1DA,mix=∑j≠Ayj′DAj\dfrac{1}{D_{A,mix}} = \sum_{j \ne A}\dfrac{y'_j}{D_{Aj}} (on the formula sheet; also given in the question)
  • Film theory kc=DAB/δk_c = D_{AB}/\delta; penetration theory kc=2DAB/(πte)k_c = 2\sqrt{D_{AB}/(\pi t_e)} (learn this)
  1. (a(i))
    The dimensionless Sherwood number (Sh) is a mass transfer coefficient that is function of Schmidt number (Sc). Explain the definition of Schmidt number by using concepts of the boundary layers of momentum and concentration.
    [3]Third
  2. (a(ii))
    Explain the difference between the penetration theory and the two-phase (two-film) theory of mass transfer.
    [2]Third
  3. (a(iii))
    What is the difference between the diffusivity or diffusion coefficient and the mass transfer coefficient?
    [2]Third
  4. (a(iv))
    Prove that for equimolecular counter diffusion from a sphere to a surrounding stationary, infinite medium, the Sherwood number based on the diameter of the sphere is equal to 2.
    [3]
  5. (b(i))
    Oxygen (A) is diffusing through carbon monoxide (B) under steady-state conditions while carbon monoxide is not diffusing. The total pressure is one atmosphere and the temperature is 0∘C0^\circ\mathrm{C}. The partial pressures of oxygen at the interface of the two gases and the bulk of carbon monoxide are 100 mmHg and 50 mmHg, respectively. The thickness of the interface between the two gases is 0.2 cm and the diffusivity DABD_{AB} is 0.185 cm2 s−10.185\ \mathrm{cm^2\,s^{-1}}. (Data: 1 atmosphere = 760 mmHg = 1.013×1051.013 \times 10^5 Pa.) Calculate the flux of diffusion (mol m−2 s−1\mathrm{mol\,m^{-2}\,s^{-1}}) of oxygen in carbon monoxide.
    [5]2:2
  6. (b(ii))
    If the total pressure was doubled, would the flux of mass transfer increase, decrease or remain the same?
    [5]
  7. (b(iii))
    Assuming that the non-diffusing gas (carbon monoxide) is replaced by a mixture of methane (B) and hydrogen (C) in volume ratio 2:1, re-calculate the flux of diffusion (mol m−2 s−1\mathrm{mol\,m^{-2}\,s^{-1}}) if the diffusivity of oxygen in methane is DO2−CH4=0.184 cm2 s−1D_{O_2-CH_4} = 0.184\ \mathrm{cm^2\,s^{-1}} and the diffusivity of oxygen in hydrogen is DO2−H2=0.690 cm2 s−1D_{O_2-H_2} = 0.690\ \mathrm{cm^2\,s^{-1}}. (Data: Di,mix=1/∑j≠i(yj′/Dij)D_{i,mix} = 1/\sum_{j \ne i}(y'_j/D_{ij}), yj′=yj/∑j≠iyjy'_j = y_j/\sum_{j \ne i}y_j.)
    [5]2:2